Let $\Omega $ open, smooth and bounded. Why $$K=\{u\in H^1(\Omega )\mid u|_{\partial \Omega }=g\}$$ is a hilbert space ? I recall that $H^1(\Omega )=W^{1,2}(\Omega )$.

Indeed, it's even not stable for the addition. Let $u,v\in K$, then $$(u+v)|_{\partial \Omega }=u|_{\partial \Omega }+v|_{\partial \Omega }=g+g=2g.$$ What's the problem here ?


Let $g\in H^{1/2}(\partial \Omega )$ and $f\in L^2(\Omega )$. Prove that there is a unique weak solution of $$\begin{cases} -\Delta u-u=f&\Omega \\ u=g&\partial \Omega \end{cases}.$$

The solution goes like :

We have to solve it in $$H=\{u\in H^1(\Omega )\mid u|_{\partial \Omega }=g\}.$$

The variational equation is given by $$\int_\Omega \nabla \varphi\cdot \nabla u-\int_\Omega u\varphi=\int_{\partial \Omega }g\varphi+\int f\varphi,\quad \varphi\in H.$$

We set $$a(u,\varphi)=\int_\Omega \nabla \varphi\cdot \nabla u-\int_\Omega u\varphi,$$ and $$T(\varphi)=\int_{\partial \Omega }g\varphi+\int f\varphi.$$ It's easy to prove that $a$ is continuous and coercive and $T$ is continuous. Lax-Milgram allow us to conclude.

Question : Since $H$ it's not a Hilbert space, the argument is wrong ? But it's an official solution of my course, so it should be a correct argument. By the way I also see this argument here (page 51-52 for people who read french). So I guess this argument is true... So how can it be true since $H$ is even not an Hilbert space ?

  • $\begingroup$ It is not a Hilbert space - $H_0^1(\Omega)$ is a Hilbert space. $\endgroup$ – Yiorgos S. Smyrlis Jun 17 '17 at 16:22
  • $\begingroup$ @YiorgosS.Smyrlis: So why to solve $$\begin{cases} \Delta u+u=f&\Omega \\ u=g&\partial \Omega \end{cases}$$ we can use Lax-Milgram in $\{u\in H^1\mid u|_{\partial \Omega }=g\}$ ? $\endgroup$ – MathBeginner Jun 17 '17 at 16:34

$K$ is infact not a hilbert space.Actually you do not use lax milagram directly for this problem. First you have to convert this to an equivalent homogeneous problem in $H_0^1$ and then you use lax milgram. This can be done by substracting off a function in $H^1$ with a trace equal to g. This is always possible due to trace theorem.


The argument of your teacher is wrong ! To use Lax-Milgram, you must be in a Hilbert space ! The link you put solve the problem in $H_0^1(\Omega )$. They do as following : Since the trace is surjective, there is $v\in H^1$ s.t. $v|_{\partial \Omega }=g$. Let $w=u-v$. Then your system is equivalent to $$(S):\begin{cases}\Delta w-w=f-\Delta v-v&\Omega \\ w=0&\partial \Omega .\end{cases}$$

You can apply your reasoning $(S)$ and you'll have what you want.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.